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<h3 class="heading"><span class="type">Paragraph</span></h3>
<p><dfn class="terminology">Proof</dfn> We start from</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
k_1 f(x)+k_2g(x)=0 \quad \textrm{in}~I.
\end{equation*}
</div>
<p class="continuation">We need to show that <span class="process-math">\(k_1\)</span> and <span class="process-math">\(k_2\)</span> must be zero.Differentiation gives</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
k_1 f^{\prime}(x)+k_2 g^{\prime}(x)=0.
\end{equation*}
</div>
<p class="continuation">Consider at <span class="process-math">\(x=x_0\text{:}\)</span></p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\begin{aligned}
&amp;k_1 f(x_0)+k_2 g(x_0)=0,\\
&amp;k_1 f^{\prime}(x_0)+k_2 g^{\prime}(x_0)=0.
\end{aligned}
\end{equation*}
</div>
<p class="continuation">The coefficient matrix determinant:</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\left|  \begin{array}{ll}
f(x_0) &amp; g(x_0)\\
f^{\prime}(x_0)  &amp;g^{\prime}(x_0) 
\end{array}
\right| =W(f, g)|_{x=x_0} \neq 0.
\end{equation*}
</div>
<p class="continuation">This implies <span class="process-math">\(k_1\)</span> and <span class="process-math">\(k_2\)</span> must be zero.</p>
<span class="incontext"><a href="sec3_3.html#p-88" class="internal">in-context</a></span>
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